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  Leetcode-316-去除重复的字母
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      <h1 id="Leetcode-316-去除重复字母"><a href="#Leetcode-316-去除重复字母" class="headerlink" title="Leetcode-316-去除重复字母"></a>Leetcode-316-<a href="https://leetcode-cn.com/problems/remove-duplicate-letters/" target="_blank" rel="noopener">去除重复字母</a></h1><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><ul>
<li>给你一个<strong>仅包含小写字母</strong>的字符串，请你去除字符串中重复的字母，使得每个字母只出现一次。</li>
<li><strong>保证返回结果的字典序最小（要求不能打乱其他字符的相对位置）。</strong></li>
</ul>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">示例 1:</span><br><span class="line"></span><br><span class="line">输入: &quot;bcabc&quot;</span><br><span class="line">输出: &quot;abc&quot;</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">示例 2:</span><br><span class="line"></span><br><span class="line">输入: &quot;cbacdcbc&quot;</span><br><span class="line">输出: &quot;acdb&quot;</span><br></pre></td></tr></table></figure>

<a id="more"></a>



<h2 id="方法：栈-哨兵"><a href="#方法：栈-哨兵" class="headerlink" title="方法：栈+哨兵"></a>方法：栈+哨兵</h2><p>思路分析：</p>
<ul>
<li><strong>首先解释一下什么是字典序。</strong><ul>
<li>字典序是指从前到后比较两个字符串的大小</li>
<li>首先比较第一个字符，如果不同则第一个字符较小的字符串更小</li>
<li>如果相同则继续比较第二个字符…. 如果比到最后都一样的话，那么两个字符串相等</li>
</ul>
</li>
</ul>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line">观察示例 1：bcabc。</span><br><span class="line"></span><br><span class="line">字符 a 在字符串中只出现一次，根据题目要求，字符 a 必须被选取；</span><br><span class="line">字符 b 出现了两次，显然选择 a后面的那个，因为字典序 ab 在 ba 前面。同理，有两个相同的字符 c ，我们选择后一个。因此，输出就是 abc。</span><br></pre></td></tr></table></figure>



<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">再观察示例 2：cbacdcbc。</span><br><span class="line"></span><br><span class="line">有 4 个字符：</span><br><span class="line">a、b、c、d。其中 a 和 d 只出现一次，必须被选取；</span><br><span class="line">b 出现 2 次，一个在 a 前面，一个在 a 后面，显然保留在 a 后面的；</span><br><span class="line">c 出现 4 次，我们把几种可能都列出来观察一下：</span><br><span class="line">情况 1：cadb</span><br><span class="line">情况 2：acdb（字典序最小）</span><br><span class="line">情况 3：adcb</span><br><span class="line">情况 4：adbc</span><br></pre></td></tr></table></figure>



<ul>
<li>下面我们就要思考，当遍历到字符串<strong>ASCII 值减少的时候</strong>，应该如何处理。（一种最理想的情况是：<code>abcd</code>，在遍历的时候，遇到的字符串的 <strong>ASCII 值逐渐增大</strong>。）</li>
</ul>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line">还看示例 1：</span><br><span class="line">已经读到了 bc，已经是字典序最小的排列。</span><br><span class="line">即将读到的 a 比 c 的 ASCII 值小。如果 a 能排在 c 之前，就能得到一个比 ca 更小的字典序 ac。</span><br><span class="line">那么 a 能不能排在 c 之前，就看 a 的后面还会不会出现字符 c，显然会。同理，由于字符 b 在将来还会出现，构成的字典序更小，因此舍弃之前的字符 b。</span><br><span class="line">到此为止，应该想到我们需要借助栈帮助我们完成这题。</span><br></pre></td></tr></table></figure>



<ul>
<li>然后根据这个思路，我们再看一下示例 2：<code>cbacdcbc</code>。</li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20200629/143955298.png" alt="mark"></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line">第 1 步：读到 c，入栈，此时栈中元素 [c]；</span><br><span class="line"></span><br><span class="line">第 2 步：读到 b，b 比之前 a 小，c 在以后还会出现，因此 c 出栈，b 入栈，此时栈中元素 [b]；</span><br><span class="line"></span><br><span class="line">第 3 步：读到 a，a 比之前 b 小，b 在以后还会出现，因此 b 出栈，a 入栈，此时栈中元素 [a]；</span><br><span class="line"></span><br><span class="line">第 4 步：读到 c，c 比之前 a 大，直接让 c 入栈，此时栈中元素 [a, c]；</span><br><span class="line"></span><br><span class="line">第 5 步：读到 d，d 比之前 d 大，直接让 d 入栈，此时栈中元素 [a, c, d]；</span><br><span class="line"></span><br><span class="line">第 6 步：读到 c，这里要注意：此时栈中已经有 c 了，此时栈中元素构成的字符顺序就是最小的字典序，不可能舍弃之前的 c，而用现在的读到的 c，因此这个 c 不需要，直接跳过；</span><br><span class="line"></span><br><span class="line">第 7 步：读到 b，b 比之前的 d 小，但是，后面不会再出现 d 了，因此 b 就应该放在这个位置，因此让 b 入栈，此时栈中元素 [a, c, d, b]；</span><br><span class="line"></span><br><span class="line">第 8 步：读到 c，同第 6 步，这个 c 我们不需要。</span><br></pre></td></tr></table></figure>



<ul>
<li><p>于是：我们可以设计如下算法：</p>
<ul>
<li>遍历字符串里的字符，<strong>如果读到的字符的 ASCII 值是升序</strong>，依次存到一个栈中；</li>
<li>如果读到的栈中已经存在，那么这个字符我们不需要。</li>
<li>如果<strong>读到的ASCII 值比栈顶元素严格小，看看栈顶后面的是否还会再出现</strong>，如果还会出现，则舍弃栈顶元素，而选择后面出现的那个元素，这样得到的字典序更小。</li>
</ul>
</li>
<li><p>因为需要判断读到的字符在栈中是否已经存在，因此可以使用哈希表，又因为题目说，字符只会出现小写字母，<strong>用一个布尔数组也是可以的，注意在出栈入栈的时候，需要同步更新一下这个布尔数组。</strong></p>
</li>
<li><p>又因为要<strong>判断栈顶元素在后面是否会被遍历到，因此我们需要首先遍历一次字符，存一下这个字符最后出现的位置，就能判断栈顶元素在后面是否会被遍历到</strong></p>
</li>
</ul>
<p><strong>1. 未使用哨兵的版本</strong></p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> String <span class="title">removeDuplicateLetters</span><span class="params">(String s)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> len = s.length();</span><br><span class="line">        <span class="comment">// 1. 特判</span></span><br><span class="line">        <span class="keyword">if</span>(len &lt; <span class="number">2</span>)&#123;                            <span class="comment">// 这里说明长度为0或者1，直接添加即可</span></span><br><span class="line">            <span class="keyword">return</span> s;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 2. 记录每个字符串已经出现的字符,并且记录每个字符出现的最后一个位置</span></span><br><span class="line">        <span class="keyword">boolean</span>[] set = <span class="keyword">new</span> <span class="keyword">boolean</span>[<span class="number">26</span>];</span><br><span class="line">        <span class="keyword">int</span>[] lastIndex = <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">26</span>];</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>;i &lt; len;++i)&#123;</span><br><span class="line">            lastIndex[s.charAt(i) - <span class="string">'a'</span>] = i;   <span class="comment">// 记录每个字符串出现的最后index</span></span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">        <span class="comment">// 3. 用栈做判断</span></span><br><span class="line">        Stack&lt;Character&gt; stack = <span class="keyword">new</span> Stack&lt;&gt;();</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>;i &lt; len;++i)&#123;</span><br><span class="line">            <span class="keyword">char</span> curr = s.charAt(i);</span><br><span class="line">            <span class="keyword">if</span>(set[curr - <span class="string">'a'</span>])&#123;                <span class="comment">// 如果这个字符栈里面已经存在</span></span><br><span class="line">                <span class="keyword">continue</span>;</span><br><span class="line">            &#125;</span><br><span class="line"></span><br><span class="line">            <span class="comment">// 3.1 对入栈元素进行判断</span></span><br><span class="line">            <span class="comment">// lastIndex[stack.peek() - 'a'] &gt;= i 说明该字符后面还会再次出现</span></span><br><span class="line">            <span class="comment">// stack.peek() &gt; curr 说明栈顶字符比当前入栈元素的字典序大</span></span><br><span class="line">            <span class="keyword">while</span>(!stack.isEmpty() &amp;&amp; stack.peek() &gt; curr &amp;&amp; lastIndex[stack.peek() - <span class="string">'a'</span>] &gt; i)&#123;</span><br><span class="line">                <span class="keyword">char</span> top = stack.pop();</span><br><span class="line">                set[top - <span class="string">'a'</span>] = <span class="keyword">false</span>;</span><br><span class="line">            &#125;</span><br><span class="line"></span><br><span class="line">            <span class="comment">// 3.2 不满足上述while 循环条件: 直接加入到栈中就好</span></span><br><span class="line">            stack.push(curr);</span><br><span class="line">            set[curr - <span class="string">'a'</span>] = <span class="keyword">true</span>;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 4. 栈顶元素一次出现并且返回字符串形式</span></span><br><span class="line">        StringBuilder sb = <span class="keyword">new</span> StringBuilder();</span><br><span class="line">        <span class="keyword">while</span>(!stack.isEmpty())&#123;</span><br><span class="line">            sb.insert(<span class="number">0</span>,stack.pop());</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> sb.toString();</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>





<p><strong>使用哨兵原因</strong>：里面判断语句太长，每一次都要判断栈是否为空。这里使用一个哨兵的技巧，一开始就在栈里面放一个最小字符’a’ 。因为后面判断语句<code>stack.peek() &gt; currentChar</code> 这里是严格符号，因此这个 <code>a</code> 一定不会被弹出。</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">while</span> (!stack.empty() &amp;&amp; stack.peek() &gt; currentChar &amp;&amp; lastAppearIndex[stack.peek() - <span class="string">'a'</span>] &gt;= i) &#123;</span><br><span class="line">    <span class="keyword">char</span> top = stack.pop();</span><br><span class="line">    set[top - <span class="string">'a'</span>] = <span class="keyword">false</span>;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>







<p><strong>使用哨兵：</strong></p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> String <span class="title">removeDuplicateLetters</span><span class="params">(String s)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> len = s.length();</span><br><span class="line">        <span class="comment">// 1. 特判</span></span><br><span class="line">        <span class="keyword">if</span>(len &lt; <span class="number">2</span>)&#123;</span><br><span class="line">            <span class="keyword">return</span> s;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 2.1 记录每个字符出现的最后一个位置，并且记录当前的字符是否已经出现过</span></span><br><span class="line">        <span class="keyword">int</span>[] lastIndex = <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">26</span>];</span><br><span class="line">        <span class="keyword">boolean</span>[] set  = <span class="keyword">new</span> <span class="keyword">boolean</span>[<span class="number">26</span>];                   <span class="comment">// 用于记录当前的字符是否已经出现过</span></span><br><span class="line">        <span class="keyword">char</span>[] chars = s.toCharArray();</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>;i &lt; len;i++)&#123;</span><br><span class="line">            lastIndex[chars[i] - <span class="string">'a'</span>] = i;                  <span class="comment">// 记录每个字符出现的最后一个位置</span></span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 2.2 使用哨兵 ： 栈初始化的时候就加入哨兵 : 字符'a'</span></span><br><span class="line">        LinkedList&lt;Character&gt; stack = <span class="keyword">new</span> LinkedList&lt;&gt;();</span><br><span class="line">        stack.addLast(<span class="string">'a'</span>);</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 3. 栈做判断</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> i = <span class="number">0</span>;i &lt; len;i++)&#123;</span><br><span class="line">            <span class="keyword">char</span> curr = chars[i];</span><br><span class="line">            <span class="keyword">if</span>(set[curr - <span class="string">'a'</span>])&#123;                            <span class="comment">// 说明当前字符之前已经出现过</span></span><br><span class="line">                <span class="keyword">continue</span>;</span><br><span class="line">            &#125;</span><br><span class="line"></span><br><span class="line">            <span class="comment">// 3.1 栈顶元素比当前元素大 且 这个字符后面还会再次出现</span></span><br><span class="line">            <span class="keyword">while</span>(stack.getLast() &gt; curr &amp;&amp; lastIndex[stack.getLast() - <span class="string">'a'</span>] &gt; i)&#123;</span><br><span class="line">                <span class="keyword">char</span> top = stack.removeLast();              <span class="comment">// 弹出栈顶元素</span></span><br><span class="line">                set[top - <span class="string">'a'</span>] = <span class="keyword">false</span>;                     <span class="comment">// 将字符改为出现过</span></span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 3.2 不满足while循环条件 ： 说明当前字符字典序是升序的 或者 当前字符之前没有出现过</span></span><br><span class="line">            stack.addLast(curr);</span><br><span class="line">            set[curr - <span class="string">'a'</span>] = <span class="keyword">true</span>;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 4. 转为字符串返回结果</span></span><br><span class="line">        StringBuilder sb = <span class="keyword">new</span> StringBuilder();</span><br><span class="line">        stack.removeFirst();</span><br><span class="line">        <span class="keyword">while</span>(!stack.isEmpty())&#123;</span><br><span class="line">            sb.append(stack.removeFirst());</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> sb.toString();</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>



<p><strong>复杂度分析：</strong></p>
<ul>
<li>时间复杂度：O(N)，这里 N 是字符的长度；</li>
<li>空间复杂度：O(N) ，最坏情况下，这个字符串本身就是字典序最小的字符串，栈中就要存字符串长度这么多的字符串。</li>
</ul>
<h2 id="关于栈的题目"><a href="#关于栈的题目" class="headerlink" title="关于栈的题目"></a>关于栈的题目</h2><p>「力扣」第 20 题：有效的括号（简单）<br>「力扣」第 71 题：简化路径（中等）<br>「力扣」第 150 题：逆波兰表达式求值（中等）<br>「力扣」第 32 题：最长有效括号（困难）<br>「力扣」第 739 题：每日温度（中等）<br>「力扣」第 496 题：下一个更大元素 I（简单）<br>「力扣」第 84 题：柱状图中最大的矩形（困难）<br>「力扣」第 42 题：接雨水（困难）<br>「力扣」第 901 题：股票价格跨度（中等）<br>「力扣」第 581 题：最短无序连续子数组（中等）<br>「力扣」第 402 题：移掉K位数字（中等）<br>「力扣」第 321 题：拼接最大数（困难）<br>「力扣」第 1673 题：找出最具竞争力的子序列（中等）</p>

      
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</div>

<link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/photoswipe@4.1.3/dist/photoswipe.min.css">
<link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/photoswipe@4.1.3/dist/default-skin/default-skin.css">
<script src="https://cdn.jsdelivr.net/npm/photoswipe@4.1.3/dist/photoswipe.min.js"></script>
<script src="https://cdn.jsdelivr.net/npm/photoswipe@4.1.3/dist/photoswipe-ui-default.min.js"></script>

<script>
    function viewer_init() {
        let pswpElement = document.querySelectorAll('.pswp')[0];
        let $imgArr = document.querySelectorAll(('.article-entry img:not(.reward-img)'))

        $imgArr.forEach(($em, i) => {
            $em.onclick = () => {
                // slider展开状态
                // todo: 这样不好，后面改成状态
                if (document.querySelector('.left-col.show')) return
                let items = []
                $imgArr.forEach(($em2, i2) => {
                    let img = $em2.getAttribute('data-idx', i2)
                    let src = $em2.getAttribute('data-target') || $em2.getAttribute('src')
                    let title = $em2.getAttribute('alt')
                    // 获得原图尺寸
                    const image = new Image()
                    image.src = src
                    items.push({
                        src: src,
                        w: image.width || $em2.width,
                        h: image.height || $em2.height,
                        title: title
                    })
                })
                var gallery = new PhotoSwipe(pswpElement, PhotoSwipeUI_Default, items, {
                    index: parseInt(i)
                });
                gallery.init()
            }
        })
    }
    viewer_init()
</script>



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